Answer:
Calculate the first differences between the y-values:
[tex]\sf 3 \underset{+1}{\longrightarrow} 4 \underset{+3}{\longrightarrow} 7 \underset{+5}{\longrightarrow} 12 \underset{+7}{\longrightarrow} 19[/tex]
As the first differences are not the same, we need to calculate the second differences:
[tex]\sf 1 \underset{+2}{\longrightarrow} 3 \underset{+2}{\longrightarrow} 5 \underset{+2}{\longrightarrow} 7[/tex]
As the second differences are the same, the relationship between the variable is quadratic and will contain an [tex]x^2[/tex] term.
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To determine the quadratic equation
The coefficient of [tex]x^2[/tex] is always half of the second difference.
As the second difference is 2, and half of 2 is 1, the coefficient of [tex]x^2[/tex] is 1.
The standard form of a quadratic equation is: [tex]y=ax^2+bx+c[/tex]
(where a, b and c are constants to be found).
We have already determined that the coefficient of [tex]x^2[/tex] is 1.
Therefore, a = 1
From the given table, when [tex]x=0[/tex], [tex]y=12[/tex].
[tex]\implies a(0)^2+b(0)+c=12[/tex]
[tex]\implies c=12[/tex]
Finally, to find b, substitute the found values of a and c into the equation, then substitute one of the ordered pairs from the given table:
[tex]\begin{aligned}\implies x^2+bx+12 & = y\\ \textsf{at }(1,19) \implies (1)^2+b(1)+12 & = 19\\ 1+b+12 & = 19\\b+13 & =19\\b&=6\end{aligned}[/tex]
Therefore, the quadratic equation for the given ordered pairs is:
[tex]y=x^2+6x+12[/tex]
